1 Introduction and preliminaries
In this paper, let \(B(H)\) denote the algebra of all bounded linear operators on a complex separable Hilbert space, \(B(H)_{+}\) denote the cone of positive operators and \(K(H)\) denote the ideal of compact operators in \(B(H)\). And \((I,|\!|\!|\cdot|\!|\!|)\) is a two-sided ideal of \(B(H)\) equipped with a unitarily invariant norm \(|\!|\!|\cdot|\!|\!|\). We shall denote this by I instead of \((I,|\!|\!|\cdot|\!|\!|)\) for convenience.
For any compact operator \(A\in K(H)\), let \(S_{1}(A), S_{2}(A), \ldots\) be the eigenvalues of \(|A|=(A^{*}A)^{\frac{1}{2}}\) arranged in decreasing order. If \(A\in M_{n}\), which is the algebra of all \(n\times n\) matrices over C, we take \(S_{k}(A)=0\) for \(k>n\). A unitarily invariant norm in \(K(H)\) is a map \(|\!|\!|\cdot|\!|\!|: K(H)\longrightarrow[0,\infty]\) given by \(|\!|\!| A|\!|\!| =g(S(A))\), \(A\in K(H)\), where g is a symmetric gauge function, cf. [6, 9]. The Schatten p-norms \(\| A\|_{p}=(\sum_{j}S_{j}^{p}(A))^{\frac{1}{p}}\) for \(p\geq1\) are significant examples of the unitarily invariant norm, and \(\|\cdot\|_{2}\) is a special unitarily invariant norm which is the Hilbert–Schmidt norm defined, for \(A\in M_{n}\), as follows:
$$\Vert A \Vert _{2}^{2}=\sum _{i,j} \vert a_{ij} \vert ^{2}= \operatorname{tr}\bigl(A^{*}A\bigr). $$
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It is well known that the arithmetic–geometric mean inequality
for positive numbers a and b has been generalized in various directions.
$$\sqrt{ab}\leq\frac{a+b}{2} $$
Bhatia et al. [2] have obtained the result that if A, B, and X are \(n\times n\) matrices with A and B are positive definite, then
$$\bigl|\!\bigl|\!\bigl|A^{\frac{1}{2}}XB^{\frac{1}{2}}\bigr|\!\bigr|\!\bigr|\leq\frac {1}{2} |\!|\!| AX+XB|\!|\!|. $$
The Heinz means
is an interpolation between the arithmetic and geometric means for \(a,b\geq0\) and \(v\in[0,1]\). It is easy to see that \(H_{v}\) is symmetric and convex function for \(v\in[0,1]\) and attains its minimum at \(v=\frac {1}{2}\); thus we have
$$H_{v}(a,b)=\frac{a^{1-v}b^{v}+a^{v}b^{1-v}}{2} $$
$$\sqrt{ab}\leq H_{v}(a,b)\leq\frac{a+b}{2}. $$
The matrix version has been proved in [2]: if A, B and X are positive definite, then for every unitarily invariant norm the function
is convex on \([0,1]\), and attains its minimum at \(v=\frac{1}{2}\). Thus we have
$$g(v)=\bigl|\!\bigl|\!\bigl|A^{v}XB^{1-v}+A^{1-v}XB^{v} \bigr|\!\bigr|\!\bigr|$$
$$2\bigl|\!\bigl|\!\bigl|A^{\frac{1}{2}}XB^{\frac{1}{2}}\bigr|\!\bigr|\!\bigr|\leq \bigl|\!\bigl|\!\bigl|A^{v}XB^{1-v}+A^{1-v}XB^{v}\bigr|\!\bigr|\!\bigr|\leq |\!|\!| AX+XB|\!|\!|. $$
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The Heron means is defined by
for \(0\leq\alpha\leq1\), see [1]. This family is the linear interpolation between the geometric and the arithmetic mean. Clearly, \(F_{\alpha}\leq F_{\beta}\) whenever \(\alpha \leq\beta\).
$$F_{\alpha}(a,b)=(1-\alpha)\sqrt{ab}+\alpha\frac{a+b}{2} $$
Bhatia and Davis have proved that the inequality
is always true for \(0\leq\alpha\leq\beta\leq1\), \(\beta\geq\frac{1}{2}\), and this restriction on β is necessary in [2].
$$\biggl|\!\biggl|\!\biggl|(1-\alpha)A^{\frac{1}{2}}XB^{\frac{1}{2}}+\alpha\biggl( \frac {AX+XB}{2}\biggr)\biggr|\!\biggr|\!\biggr|\leq \biggl|\!\biggl|\!\biggl|(1-\beta)A^{\frac{1}{2}}XB^{\frac {1}{2}}+ \beta\biggl(\frac{AX+XB}{2}\biggr)\biggr|\!\biggr|\!\biggr|$$
The Heinz means and the Heron means satisfy the inequality
for \(0\leq v\leq1\). And \(\alpha(v)=1-4(v-v^{2})\), this is a convex function, its minimum value is \(\alpha(\frac{1}{2})=0\), and its maximum value is \(\alpha(0)=\alpha(1)=1\).
$$H_{v}(a,b)\leq F_{\alpha(v)}(a,b) $$
In [10] the authors have presented the result that
for \(\frac{1}{4}\leq v\leq\frac{3}{4}\) and \(\alpha\in[\frac{1}{2},\infty ]\) and they used the properties of contractive map on I to prove it.
$$\frac{1}{2}\bigl|\!\bigl|\!\bigl|A^{v}XB^{1-v}+A^{1-v}XB^{v} \bigr|\!\bigr|\!\bigr|\leq \biggl|\!\biggl|\!\biggl|(1-\alpha)A^{\frac{1}{2}}XB^{\frac{1}{2}}+\alpha \biggl(\frac {AX+XB}{2}\biggr) \biggr|\!\biggr|\!\biggr|$$
A different version of the Heinz inequality,
for \(\alpha\in[0,1]\) was proved by Bhatia and Davis [3] in 1995.
$$\bigl|\!\bigl|\!\bigl|A^{\alpha}XB^{1-\alpha}-A^{1-\alpha}XB^{\alpha} \bigr|\!\bigr|\!\bigr|\leq|2\alpha-1||\!|\!| AX-XB|\!|\!|, $$
A further generalization, namely
was proved for \(t\in[-2,2]\), and \(\alpha\in[\frac{1}{4},\frac{3}{4}]\). For more details refer to [13] and [14].
$$\frac{2+t}{2}\bigl|\!\bigl|\!\bigl|A^{\alpha}XB^{1-\alpha}+A^{1-\alpha}XB^{\alpha } \bigr|\!\bigr|\!\bigr|\leq \bigl|\!\bigl|\!\bigl|AX+XB+tA^{\frac{1}{2}}XB^{\frac{1}{2}}\bigr|\!\bigr|\!\bigr|, $$
Contractive maps on I play a key role to prove the inequalities, more details refer to [4, 5, 7, 8, 11]. This in turn proves the corresponding Schur multiplier maps to be contractive. We use the ideals to establish the fact that some special maps on I are contractive.
Our paper consists of three parts. In the second part, we will give a new norm inequality which is proved by the properties of contractive maps on I. In the third part, we will present a inequality related to the Heinz means and we notice that this inequality is a comparison between the Heinz means and the geometric mean. Our results are extensions of some previous conclusions about norm inequalities.
2 Norm inequalities with contractive maps
Let \(L_{X}\), \(R_{Y}\) denote the left and right multiplication maps on \(B(H)\), respectively, that is, \(L_{X}(T)=XT\) and \(R_{Y}(T)=TY\) and we have
$$ e^{L_{X}+R_{Y}}(T)=e^{X}Te^{Y}. $$
(1)
Neeb [12] has proved that \((L_{X}\pm R_{Y})^{-1}\sin(L_{X}\pm R_{Y})\) is an expansive map on I by using the Weierstrass factorization theorem. We use \(X_{1}\) and \(Y_{1}\) to denote two selfadjoint operators in \(B(H)\) and \(D=L_{X_{1}}-R_{Y_{1}}\). The following proposition is a result for a contractive map in I; for more details refer to [10].
Proposition 2.1
([10, Proposition 2.4])
Let
\(0\leq r\leq s\)
or
\(s\leq r\leq0\), then each of the following operator maps is a contraction on
I.
(1)
\(\frac{(1+t)\cosh(\frac{r}{2}D)}{t+\cosh(sD)}\)
for
\(|t|\leq1\).
(2)
\(\frac{t+\cosh(rD)}{t+\cosh(sD)}\)
for
\(|t|\leq1\).
(3)
\(\frac{\cosh(rD)}{\cosh(sD)}\). In particular
\(\frac{1}{r}\int\frac {\cosh(rD)}{\cosh(sD)}\,dr=\frac{\sinh(rD)}{rD\cosh(sD)}\)
is a contraction.
(4)
\(\frac{s\sinh(rD)}{r\sinh(sD)}\). The case
\(r=0\)
would become
\(\frac {sD}{\sinh(sD)}\).
Corollary 2.2
([10, Corollary 2.5])
The following maps are contractive on
I.
(1)
\(\frac{(t+1)\cosh((2v-1)D)}{t+\cosh D}\)
for
\(\frac{1}{4}\leq v\leq \frac{3}{4}\), and
\(| t|\leq1\).
(2)
\(\frac{2\cosh(rD)}{\cosh(s_{1}D)+\cosh(s_{2}D)}\)
for
\(0\leq r\leq \max\{ \frac{s_{1}+s_{2}}{2},\frac{s_{1}-s_{2}}{2}\}\)
or
\(\min \{ \frac {s_{1}+s_{2}}{2},\frac{s_{1}-s_{2}}{2}\}\leq r\leq0\).
(3)
\(\frac{(s_{1}+s_{2})\sinh(rD)}{r(\sinh(s_{1}D)+\sinh(s_{2}D))}\)
for
\(0\leq r\leq\frac{s_{1}+s_{2}}{2}\)
or
\(\frac{s_{1}+s_{2}}{2}\leq r\leq 0\).
Corollary 2.3
Let
\(0\leq v\leq1\)
and
\(|t|\leq1\). Then the map
\(\frac{(t+1)\cosh ((2v-1)D)}{t+\cosh(2D)}\)
is contractive on
I.
Proof
This map is a special case of \((1)\) of Proposition 2.1 when \(|s|=2\). From the condition \(0\leq v\leq1\), we can obtain \(| 2v-1|\leq1\). Letting \(\frac{r}{2}=2v-1\), we get the desired result. □
With these above results about contractive maps in I, we obtain a new norm inequality which derives from the Heinz means and other related inequalities. Let R be an invertible operator in \(B(H)_{+}\), then there exists a selfadjoint operator \(S\in B(H)\) such that \(R=e^{S}\). To avoid repetitions, we denote the two invertible operators A and B in \(B(H)_{+}\) by \(e^{2X_{1}}\) and \(e^{2Y_{1}}\), respectively, where \(X_{1}\) and \(Y_{1}\) in \(B(H)\) are selfadjoint. The corresponding operator map \(L_{X_{1}}-R_{Y_{1}}\) is denoted by D.
Theorem 2.4
Let
\(v\in[0,1]\)
and
\(\alpha\in[\frac{1}{2},\infty)\). Supposed that
A
and
B
are any two invertible operators in
\(B(H)_{+}\), \(X\in I\), then
$$ \frac{1}{2}\bigl|\!\bigl|\!\bigl|A^{v}XB^{1-v}+A^{1-v}XB^{v} \bigr|\!\bigr|\!\bigr|\leq \biggl|\!\biggl|\!\biggl|(1-\alpha)A^{\frac{1}{2}}XB^{\frac{1}{2}}+\alpha \frac{A^{\frac {3}{2}}XB^{-\frac{1}{2}}+A^{-\frac{1}{2}}XB^{\frac{3}{2}}}{2}\biggr|\!\biggr|\!\biggr|. $$
(2)
Proof
Put \(A^{\frac{1}{2}}XB^{\frac{1}{2}}=T\) and use (1) we can notice that
And we also have
Setting \(\frac{1}{1+t}=\alpha\) in (4) and applying Corollary 2.3, we can obtain (2) from (3) and (4). □
$$\begin{aligned} &\frac{A^{v}XB^{1-v}+A^{1-v}XB^{v}}{2} \\ &\quad =\frac{A^{\frac{2v-1}{2}}A^{\frac{1}{2}}XB^{\frac{1}{2}}B^{\frac {1-2v}{2}}+A^{\frac{1-2v}{2}}A^{\frac{1}{2}}XB^{\frac{1}{2}}B^{\frac {2v-1}{2}}}{2} \\ &\quad =\frac{A^{\frac{2v-1}{2}}TB^{\frac{1-2v}{2}}+A^{\frac {1-2v}{2}}TB^{\frac{2v-1}{2}}}{2} \\ &\quad =\frac{e^{(2v-1)(L_{X_{1}}-R_{Y_{1}})}T+e^{-(2v-1)(L_{X_{1}}-R_{Y_{1}})}T}{2} \\ &\quad =\cosh\bigl((2v-1)D\bigr)T. \end{aligned}$$
(3)
$$\begin{aligned} &(1-\alpha)A^{\frac{1}{2}}XB^{\frac{1}{2}}+\alpha \frac{A^{\frac {3}{2}}XB^{-\frac{1}{2}}+A^{-\frac{1}{2}}XB^{\frac{3}{2}}}{2} \\ &\quad =(1-\alpha)T+\alpha\frac{AA^{\frac{1}{2}}XB^{\frac {1}{2}}B^{-1}+A^{-1}A^{\frac{1}{2}}XB^{\frac{1}{2}}B}{2} \\ &\quad =(1-\alpha)T+\alpha\frac{ATB^{-1}+A^{-1}TB}{2} \\ &\quad =(1-\alpha)T+\alpha\frac {e^{2(L_{X_{1}}-R_{Y_{1}})}T+e^{-2(L_{X_{1}}-R_{Y_{1}})}T}{2} \\ &\quad =\bigl[(1-\alpha)+\alpha\cosh(2D)\bigr]T. \end{aligned}$$
(4)
3 Norm inequality related to the Heinz means
According to the above results, we set
$$\begin{aligned} &H_{v}(a,b)=\frac{a^{v}b^{1-v}+a^{1-v}b^{v}}{2}, \\ &G_{v}(a,b)=(1-v)a^{\frac{1}{2}}b^{\frac{1}{2}}+v\frac{a^{\frac {3}{2}}b^{-\frac{1}{2}}+a^{-\frac{1}{2}}b^{\frac{3}{2}}}{2}. \end{aligned}$$
Then we have the following result, which is a interpolation of \(H_{v}\) and \(G_{v}\).
Theorem 3.1
Let
\(a,b>0\)
and
\(v\in[\frac{1}{2},1]\). Then
$$ H_{v}(a,b)\leq\biggl(\frac{H_{\frac{1}{2}}(a,b)}{H_{1}(a,b)}\biggr)^{1-v}G_{v}(a,b). $$
(5)
Proof
Without loss of generality, we can suppose that \(a=1\). Then the inequality (5) can be simplified to
To prove this inequality, let
Calculations show that
We can notice that \(f^{\prime\prime}(v)\geq0\) for \(\frac{1}{2}\leq v\leq1\), thus f is convex on \([\frac{1}{2},1]\). Then we have \(f(v)\leq\max\{f(\frac{1}{2}),f(1)\}\). By calculation, we have
because of \(\frac{\sqrt{2}b^{\frac{1}{4}}(1+b)^{\frac{1}{2}}}{b^{\frac {1}{2}}+\frac{1}{2}b^{-\frac{1}{2}}+\frac{1}{2}b^{\frac{3}{2}}}\leq 1\) (\(b>0\)), which equals \(\sqrt{2}b^{\frac{1}{4}}(1+b)^{\frac{1}{2}}\leq b^{\frac{1}{2}}+\frac{1}{2}b^{-\frac{1}{2}}+\frac{1}{2}b^{\frac {3}{2}}\), that is, \(8(1+b)b^{\frac{3}{2}}\leq6b^{2}+b^{4}+4b^{3}+4b+1\).
$$\frac{b^{v}+b^{1-v}}{2}\leq\biggl(\frac{2\sqrt{b}}{1+b}\biggr)^{1-v}\biggl[(1-v) \sqrt {b}+v\frac{b^{-\frac{1}{2}}+b^{\frac{3}{2}}}{2}\biggr]. $$
$$f(v)=\log\bigl(b^{v}+b^{1-v}\bigr)-(1-v)\log \frac{2\sqrt{b}}{1+b}-\log\bigl[2(1-v)\sqrt {b}+v\bigl(b^{-\frac{1}{2}}+b^{\frac{3}{2}} \bigr)\bigr]. $$
$$f^{\prime\prime}(v)=\frac{4b^{v}b^{1-v}\log ^{2}b}{(b^{v}+b^{1-v})^{2}}+\frac{(b^{-\frac{1}{2}}+b^{\frac {3}{2}}-2b^{\frac{1}{2}})^{2}}{[(1-v)b^{\frac{1}{2}}+v(b^{-\frac {1}{2}}+b^{\frac{3}{2}})]^{2}}. $$
$$f\biggl(\frac{1}{2}\biggr)=\log\biggl(\frac{\sqrt{2}b^{\frac{1}{4}}(1+b)^{\frac {1}{2}}}{b^{\frac{1}{2}}+\frac{1}{2}b^{-\frac{1}{2}}+\frac{1}{2}b^{\frac {3}{2}}}\biggr)\leq0 $$
Setting \(b=x^{2}\) (\(x>0\)), thus we have \(x^{8}+4x^{6}-8x^{5}+6x^{4}-8x^{3}+4x^{2}+1\geq0\), that is, \((x-1)^{2}(x^{6}+2x^{5}+7x^{4}+4x^{3}+7x^{2}+2x+1)\geq0\), which is always true when \(x>0\), then we get the desired results.
In the same way,
because of \(\frac{1+b}{b^{-\frac{1}{2}}+b^{\frac{3}{2}}}\leq1\) (\(b>0\)), which equals \(1+b\leq b^{-\frac{1}{2}} +b^{\frac{3}{2}}\), that is \((b^{\frac {1}{2}}-1)(b^{\frac{3}{2}}-1)\geq0\), which is always true when \(b>0\). Hence the values of \(f(\frac{1}{2})\) and \(f(1)\) are both less than 0, that is, \(f(v)\leq0\). Then we complete the proof. □
$$f(1)=\log\biggl(\frac{1+b}{b^{-\frac{1}{2}}+b^{\frac{3}{2}}}\biggr)\leq0 $$
Remark 3.2
The inequality which we obtained from Theorem 3.1 can also be written as
and we notice that it is a new comparison between the Heinz means and the geometric means.
$$H_{v}(a,b)\leq\biggl(\frac{2\sqrt{ab}}{a+b}\biggr)^{1-v}G_{v}, $$
Corollary 3.3
Let
\(A, B\in M_{n}^{+}\), \(X\in M_{n}\)
and
\(\frac{1}{2}\leq v\leq1\). If there are two positive numbers
m, M
such that
\(m\leq A, B\leq M\), then
$$\biggl\Vert \frac{A^{v}XB^{1-v}+A^{1-v}XB^{v}}{2} \biggr\Vert _{2}\leq\biggl( \frac {m+M}{2\sqrt{mM}}\biggr)^{v-1} \biggl\Vert (1-v)A^{\frac{1}{2}}XB^{\frac {1}{2}}+v \frac{A^{\frac{3}{2}}XB^{-\frac{1}{2}}+A^{-\frac {1}{2}}XB^{\frac{3}{2}}}{2} \biggr\Vert _{2}. $$
Proof
Let \(A=U\operatorname{diag}(\lambda_{i})U^{*}\) and \(B=V\operatorname {diag}(\mu_{j})V^{*}\) be the spectral decompositions of A and B, respectively. Letting \(U^{*}XV=Y\), we have
where ∘ denotes the Schur product. Applying (5) we have
where we have used the fact that \(m\leq\lambda_{i}\), \(\mu_{j}\leq M\) to obtain the last inequality. □
$$\begin{aligned} &\frac{A^{v}XB^{1-v}+A^{1-v}XB^{v}}{2} \\ &\quad =U\frac{\operatorname{diag}(\lambda_{i}^{v})Y\operatorname{diag}(\mu _{j}^{1-v})+\operatorname{diag}(\lambda_{i}^{1-v})Y\operatorname {diag}(\mu_{j}^{v})}{2}V^{*} \\ &\quad =U\frac{[\lambda_{i}^{v}\mu_{j}^{1-v}+\lambda_{i}^{1-v}\mu _{j}^{v}]\circ[y_{ij}]}{2}V^{*}, \end{aligned}$$
(6)
$$\begin{aligned} & \biggl\Vert \frac{A^{v}XB^{1-v}+A^{1-v}XB^{v}}{2} \biggr\Vert _{2}^{2} \\ &\quad =\sum_{i,j}\biggl(\frac{\lambda_{i}^{v}\mu_{j}^{1-v}+\lambda_{i}^{1-v}\mu _{j}^{v}}{2} \biggr)^{2}| y_{ij}|^{2} \\ &\quad \leq \sum_{i,j}\biggl(\frac{\lambda_{i}+\mu_{j}}{2\sqrt{\lambda_{i}\lambda _{j}}} \biggr)^{2(v-1)}\biggl((1-v)\lambda_{i}^{\frac{1}{2}} \mu_{j}^{\frac {1}{2}}+v\frac{\lambda_{i}^{\frac{3}{2}}\mu_{j}^{-\frac{1}{2}}+\lambda _{i}^{-\frac{1}{2}}\mu_{j}^{\frac{3}{2}}}{2}\biggr)| y_{ij} |^{2} \\ &\quad \leq \biggl(\frac{m+M}{2\sqrt{mM}}\biggr)^{2(v-1)} \biggl\Vert (1-v)A^{\frac {1}{2}}XB^{\frac{1}{2}}+v\frac{A^{\frac{3}{2}}XB^{-\frac {1}{2}}+A^{-\frac{1}{2}}XB^{\frac{3}{2}}}{2} \biggr\Vert _{2}^{2}, \end{aligned}$$
(7)
Competing interests
The authors declare that they have no competing interests.
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